DISTRIBUSI BINOMIAL

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Jadi f(x) =µ = npµ = 5p2,84 = 5p5 p x qx 5x2,845p == 0,57q = 1 - p= 1 - 0,57= 0,43µ = x =µ =0 = fi xi f2 14 1 20 2 34 3 22 4 8 5100 284100= 2,84

LAPORAN KULIAH KERJA NYATA/ MAGANG - Universitas Brawijaya

rencana program & kegiatan pembelajaran semester (rpkps)

Bila 5 mata uang dilemparkan sebanyak 100kali, terdapat distribusi keluar gambar sbb :xi012345fi2142034228• Bila xi = 0 berarti selama 100 kali pelemparan 5 mata uangtidak pernah keluar gambar sebanyak 2 kali.• xi = 1 berarti selama 100 kali pelemparan 1 gambar keluarsebanyak 14 kali.• dst.

Jadi f(x) =µ = npµ = 5p2,84 = 5p5 p x qx 5x2,845p == 0,57q = 1 - p= 1 - 0,57= 0,43µ = x =µ =0 = fi xi f2 14 1 20 2 34 3 22 4 8 5100 284100= 2,84

Page 1 and 2: DISTRIBUSI BINOMIAL
Page 4 and 5: CONTOH: Seorang mahasiswa menghadap
Page 6 and 7: JAWAB:A. P(x = 1) P=4 1 = 0,386B.
Page 8 and 9: Sebuah mesin yang memproduksi semac
Page 10: PROBABILITAS BINOMIAL KUMULATIFAdal

DISTRIBUSI BINOMIAL

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